总和索引0值匹配的列表值？

Question

I have a list of lists and I want to merge them to sum the inner index[1] values where the index[0] values match. My list looks like this:

lists = [
['Gifts', [4]],
['Gifts', [4]],
['Politics', [3]],
['Supply', [4]],
['Supply', [4]],
['Prints', [1]],
['Prints', [1]],
['Prints', [1]],
['Politics', [3]],
['Politics', [3]],
['Accounts', [2]],
['Accounts', [2]],
['Accounts', [2]],
['Features', [3]],
['Features', [2]]
]

I would therefore like the new structure to be:

new_lists = [
['Gifts', 8], ['Politics', 9], ['Supply', 8], ['Prints', 3], ['Accounts', 6], ['Features', 5]
]

How do I achieve this in Python?

Answer 1

You can use defaultdict(int) from collections module:

>>> from collections import defaultdict
>>> d = defaultdict(int)
>>> for key, value in lists:
...     d[key] += value[0]
... 
>>> dict(d)
{'Gifts': 8, 'Prints': 3, 'Accounts': 6, 'Features': 5, 'Supply': 8, 'Politics': 9}
>>> list(d.items())
[('Prints', 3), ('Features', 5), ('Supply', 8), ('Gifts', 8), ('Accounts', 6), ('Politics', 9)]

Answer 2

Or, use a Counter from collections . You can initialize it either manually by incrementing the values for every key:

from collections import Counter

c = Counter()
for i, j in lists:
    c[i] += l[j]

Or, by providing a flat list of the expanded contents, Counter will then do the counting for you:

c = Counter(sum([[i] * j[0] for i,j in lists], []))

In both cases, create the result list by using a list comprehension grabbing the contents of the counter with c.items() :

r = [[i, j] for i, j in c.items()]

Result being:

print(r) 
[['Supply', 8], ['Features', 5], ['Prints', 3],
 ['Gifts', 8], ['Accounts', 6], ['Politics', 9]]

Answer 3

Use a defaultdict :

In [52]: from collections import defaultdict

In [53]: d = defaultdict(int)

In [54]: for name, (val,) in lists:
   ....:     d[name] += val
   ....:     

In [55]: d.items()
Out[55]: dict_items([('Politics', 9), ('Supply', 8), ('Prints', 3), ('Features', 5), ('Gifts', 8), ('Accounts', 6)])